Python学习——set集合的补充

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set 是一个无序且不重复的元素集合
>>> num = {1,2,3,4,5} 

1.add()添加一个元素 
>>> num.add(6) 
>>> num 
>>> {1,2,3,4,5,6}

2.clear()清除集合中所有元素 
>>> num.clear() 
>>> num 
>>> set()

3.copy()复制一个集合 
>>> num1 = num.copy() 
>>> num1 
>>> {1,2,3,4,5}

4.difference()取得集合在一个或多个集合中不同的元素 
>>> num1 = {2,4,6,8,10} 
>>> num2 = {1,3,11,12,14} 
#返回在一个集合中不同的元素 
>>> num.difference(num1) 
>>> {1,3,5} 
#返回在多个集合中不同的元素 
>>> num.difference(num1,num2) 
>>> {5}

5.difference_update()删除当前集合中所有包含在新集合里的元素 
>>> num1 = {2,4,6,8,10} 
>>> num2 = {1,2,11,13} 
>>> num.difference_update(num1,num2) 
>>> num 
>>> {3,5}

6.discard()从集合中移除一个元素，如果元素不存在，不做任何处理 
>>> num.discard(1) 
>>> num 
>>> {2,3,4,5}

7.intersection()取交集，新建一个集合 
>>> num1 ={1,3,5,7,9} 
>>> num.intersection(num1) 
>>> {1,3,5}

8.intersection_update()取交集，修改与原来的集合 
>>> num1 = {1,3,5,7,9} 
>>> num.intersection_update(num1) 
>>> num 
>>> {1,3,5}

9.isdisjoint()如果没有交集，返回True 
>>> num2 ={6,8,10} 
>>> num.isdisjoint(num2) 
>>> True

10.pop()从集合开头移除一个元素 
>>> num.pop() 
>>> 1 
>>> num 
>>> {2,3,4,5} 
PS：如果集合为空，返回错误提示

11.symmetric_difference()差集，创建新对象 
>>> num = {1,2,3,4,5,6} 
>>> num1 = {2,3,4,6,8,9} 
>>> num.symmetric_difference(num1) 
>>> {1,5,8,9}

12.symmetric_difference_update()差集，改变原来的集合 
>>> num = {1,2,3,4,5,6} 
>>> num1 = {2,3,4,6,8,9} 
>>> num.symmetric_difference_update(num1) 
>>> num 
>>> {1,5,8,9}

13.union()并集，返回一个新集合 
>>> num ={1,2,4,6,7} 
>>> num1 ={,2,4,6,8,10,12} 
>>> num.union(num1) 
>>> {1,2,4,6,7,8,10,12}

14.update()并集，并更新该集合 
>>> num ={1,2,4,6,7} 
>>> num1 ={,2,4,6,8,10,12} 
>>> num.update(num1) 
>>> num 
>>> {1,2,4,6,7,8,10,12}

小练习：

 1 old_dict = {
 2     "#1": {‘hostname‘: ‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80},
 3     "#2": {‘hostname‘: ‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80},
 4     "#3": {‘hostname‘: ‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80}     
 5 }
 6 new_dict = {
 7     "#1": {‘hostname‘: ‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 800},
 8     "#3": {‘hostname‘: ‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80},
 9     "#4": {‘hostname‘: ‘c2‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80}     
10 }
11 old_set = set(old_dict.keys()) 
12 update_list = list(old_set.intersection(new_dict.keys()))
13 
14 new_list = [] 
15 del_list = []
16 
17 for i in new_dict.keys():     
18 　　if i not in update_list:         
19 　　　　new_list.append(i)
20 for i in old_dict.keys():     
21 　　if i not in update_list:         
22 　　　　del_list.append(i)
23 print (update_list,new_list,del_list,new_dict.keys()) 
24 print(new_dict)

View Code

结果为:
>>> [‘#1‘, ‘#3‘] [‘#4‘] [‘#2‘] dict_keys([‘#1‘, ‘#3‘, ‘#4‘])
>>>{
‘#1‘: {‘mem_capicity‘: 800, ‘hostname‘: ‘c1‘, ‘cpu_count‘: 2},
‘#3‘: {‘mem_capicity‘: 80, ‘hostname‘: ‘c1‘, ‘cpu_count‘: 2},
‘#4‘: {‘mem_capicity‘: 80, ‘hostname‘: ‘c2‘, ‘cpu_count‘: 2}
}

Python学习——set集合的补充

标签：hid   color   date()   section   包含   car   cti   alt   不重复   

原文地址：http://www.cnblogs.com/huan-ge/p/6681535.html